document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=688891>Question 688891</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How much of an alloy that is 10% copper should be mixed with 600 ounces of an alloy that is 90% copper in order to get an alloy that is 40% copper? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #425731 by </B><A HREF=/tutors/aboutme.mpl?userid=checkley79><B>checkley79(3341)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=checkley79><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=425731\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> .10X+.90*600=.40(X+600)<BR>\n" );
document.write( ".10X+540=.40X+240<BR>\n" );
document.write( ".10X-.40X=240-540<BR>\n" );
document.write( "-.30X=-300<BR>\n" );
document.write( "X=-300/-.30<BR>\n" );
document.write( "X=1000 OZ. OF 10% COPPER IS USED.<BR>\n" );
document.write( "PROOF:<BR>\n" );
document.write( ".10*1000+540=.40(1000+600)<BR>\n" );
document.write( "100+540=.40*1600<BR>\n" );
document.write( "640=640<BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );